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Conservation of energy problem help

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data
    1. Consider a rod of mass 1g and length L, held in place against a vertical wall by
    a cable at the opposite end as shown in the diagram. A sign of mass 10kg is also
    hanging o the end of the rod furthest from the wall. The angles marked are 30 degrees
    and 45 degrees .
    (a) (10 points) Find the tension in the cable. (Hint: Is the mass of the rod negligible
    here? Then clearly mark your pivot point and start summing your
    torques.) If the cable has a breaking point of 1000N, will it hold?
    (b) (10 points) Find the normal force from the wall on the rod and explain
    whether or not there must also be friction between the wall and the rod in
    this particular problem. If so, give the magnitude and direction of the friction
    force required. (Hint: you will need to draw a diagram and sum the forces.)



    2. Relevant equations



    3. The attempt at a solution
    - the rod has a mass of 1kg
    er 1g
    hm i guess its negilible
    so sign is 1kg, its 1L away from the pivot point
    torque is force times distance
    is torque caused by the rod is 1kg*1L*10m/s^2

    I'm not really sure how to figure this out, but that is how far I got
     
  2. jcsd
  3. Nov 29, 2011 #2
    diagram?
     
  4. Nov 29, 2011 #3
    here's the attached diagram
     

    Attached Files:

    • ll.png
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  5. Nov 29, 2011 #4
    Is the rod of MASS = 1 gram or has WEIGHT = 1 x g = 10N where acc due to gravity is taken as 10m/s^2?

    To find the tension it is best to find moments about the pivot so that forces at the pivot will not enter the equation.
     
  6. Nov 29, 2011 #5
    i still dont understand...:confused:
     
  7. Nov 29, 2011 #6
    Do you know how to find moment of a force about a point?
     
  8. Nov 29, 2011 #7
    I think you know.
    Moment =torque
     
  9. Nov 29, 2011 #8
    no i don't what does that mean
     
  10. Nov 29, 2011 #9
    how do find the tension the cable
     
  11. Nov 29, 2011 #10
    torque of a force about a point = force x perpendicular distance of line of action of force from the the given point.

    hence

    torque of 100N from point of attachment of cable with wall = 100 x L sin45.

    What is the torque of the weight of the rod, 10N with the same point of attachment?
     
  12. Nov 29, 2011 #11
    so if i do 1000Nsin45 will that be correct
     
  13. Nov 30, 2011 #12
    Some help so that you can try to correct your last post:

    As I told you before

    torque of a force about a point = force x perpendicular distance of line of action of force from the the given point

    therefore torque of weight of rod about point point of attachment of cable with wall =
    weight x perp dist of line of action of W with point of ...
    = 100 x ...
     
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